Question

A force F~ = Fx ˆı + Fy ˆ acts on a particle that

undergoes a displacement of ~s = sx ˆı + sy ˆ
where Fx = 3 N, Fy = −2 N, sx = 5 m, and
sy = 2 m.
Find the work done by the force on the
particle.
Answer in units of J.
Find the angle between F~ and ~s.
Answer in units of ◦

Answer 1

Step-by-step explanation:

Given:

F = Fₓ·i + Fy·j

S = Sₓ·i + Sy·j

Fₓ = 3 N

Fy = - 2 N

Sₓ = 5 m

Sy = 2 m

_________

A - ? - Work

α - ? - Angle

F = 3·i - 2·j

S = 5·i + 2·j

The work is numerically equal to the scalar product of the displacement force

A = (F·S) = 3·5 + (-2)·2 = 15 - 4 = 11 J

Modules:

| F | = √ (3² + (-2)² ) = √ (9 + 4) = √ 13

| S | = √ (5² + 2² ) = √ (25 + 4) = √ 29

Angle:

cos α = (F·S) / ( |F| · |S| ) = 11 / ( √13 · √29) ≈ 0,5665

α ≈ 55.5°